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|(1) Posted by Rajendiran Raju [Tuesday, Apr 29, 2014 07:24]|
twomover : tries ?!
Is there any twomover with the following Content.,
for example , Whites total Possible first moves is 11
in that One is Key Move, all the other 10 moves are tries !!
[Possible moves 8 (1:7) ; Possible moves 10 (1:9) Vice Versa ]
|(2) Posted by Kevin Begley [Tuesday, Apr 29, 2014 08:42]|
...including keys/tries based upon the threat of zugzwang?
|(3) Posted by Hauke Reddmann [Tuesday, Apr 29, 2014 13:13]|
A mild "yes" - at least I pondered about exactly this theme
and even composed a problem...can't find it at the moment;
I possibly never saved it.
This would surely make an interesting theme tourney!
An immediate question is which "cheats" are allowed.
I know there exists a record problem where all (?!) moves
except the solution are tries. Possibly in "Schach und Zahl"
or so. But as you can guess, the refutation is always the
same. (AFAIK it was NOT stalemate, which would ease the
construction even more.)
Here is a random attempt with stalemate (note that 1.Kd3? Kb1!):
(= 16+1 )
You can surely increase the number of tries. But I don't find
that very interesting. Much more interesting would be to increase
the number of refutations.
(= 9+5 )
Another random example - but 1.Ld6? cd,h2! would be inacceptable!
Possibly the Morse book already has the answer to that particular
question already; I look it up tomorrow.
|(4) Posted by Hauke Reddmann [Wednesday, Apr 30, 2014 11:34]|
From the "Morse already did it" dept.:
His own record is 29 (#212). At the end of that chapter,
there are problems with a) all 16 white men having a try
and b) all 16 black men having a refutation.
NONE of these problems, though, fall into the specific
demand of the original poster (namely, ALL moves except
the solution should be tries...surely in the sense that
the refutation is unique).
I thus declare this theme "MPF free-for-all" :-) Anyone want
to join? My own record of different refutations
I quickly hacked together stands at 17. You surely
can beat that one! (I post it the next days if noone tries
to beat it.)
|(5) Posted by Hauke Reddmann [Saturday, May 3, 2014 15:30]|
(= 14+9 )
D~ TxD 4
DxT KxT 1
T~ TxT 5
TxT KxT 1
La~ a1~ 1
LxS KxS 1
S~ a5 1
b5 ab 1
g4 fg 1
Le2,d3 PxL 2
18 unique refutations, 1.Kb7!
Obviously there is vast room for improvement, especially a1~ I don't
like at all (but it's acceptable, I think).
|(6) Posted by seetharaman kalyan [Saturday, May 3, 2014 17:15]|
How about adding WSa3 and BPb5? Four more tries and refutations added though losing the try 1.b5. I think it is still legal.
|(7) Posted by Geir Sune Tallaksen Østmoe [Saturday, May 3, 2014 19:38]|
Wouldn't that be illegal because White is only missing one piece, while Black's queenside pawns must have captured two? However, move bPa6 to c6 and replace wSa5 with wPc5, and it should be legal. Four added tries, at the cost of two (b5 and S~).
|(8) Posted by Geir Sune Tallaksen Østmoe [Saturday, May 3, 2014 19:43]|
By the way, is there any reason not to add wSe1 too?
|(9) Posted by seetharaman kalyan [Sunday, May 4, 2014 08:58]|
I think both your suggestions are good and will work.
|(10) Posted by Hauke Reddmann [Sunday, May 4, 2014 15:07]|
Yup, +Se1 I could have found myself. (Unfortunately, I searched
for a useful field when I still had a white Pe2 to box in the Bf1.)
OK, now we have 21. BTW, I'm just pondering about some
Nc1,Bb2,Pa3/Pa4,Pb3 setup avoiding the a1~ problem. Leaving the
Na5/Pa6 out, we are at, uhm, 20? I'm too tired to count at the moment :-)
|(11) Posted by Arno Tungler [Sunday, May 4, 2014 17:39]|
This has exactly 21 "tries" with all different refutations.
(= 13+10 )
|(12) Posted by Frank Richter [Sunday, May 4, 2014 19:52]|
Nice setting, Arno.
A possibility to add 2 tries is +wSa6, wPb7 (1.Sc7!), but it seems, that the position is not legal (due to the a-pawns ..).
|(13) Posted by Valery Liskovets [Sunday, May 4, 2014 20:46]|
Yes, Frank. But we may, instead, add wPd3 legally for one more try.
|(14) Posted by Frank Richter [Sunday, May 4, 2014 20:56]|
Yes, but that would give a doubled refutation (1.- c:d4), yes - with different pieces captured, but surely not in Arno's intention.
|(15) Posted by seetharaman kalyan [Sunday, May 4, 2014 21:06]|
Why not add Sb1? Total tries are now 23... or is it only 22?
|(16) Posted by Geir Sune Tallaksen Østmoe [Sunday, May 4, 2014 21:52]|
Adding wPb7 would make 1.Qc8 a solution, so legality is not the only issue.
Adding Nb1 would give 23 tries, but the refutation of Nc3 and Bc3 would be the same, so only 22 different refutations. I think Arno wanted to show a position where every try has a different refutation.
|(17) Posted by Geir Sune Tallaksen Østmoe [Sunday, May 4, 2014 21:55]|
Oh, wait... with Nb1, 1.Nc3 would not even be a try, since Black also has 1...Rh6!
|(18) Posted by Jacques Rotenberg [Monday, May 5, 2014 00:11]; edited by Jacques Rotenberg [14-05-05]|
The position of Arno is wonderful, it seems to me difficult to do better : all white moves are try or key, each have a specific refutation. 21 is a very big number for this.
Well done !
|(19) Posted by Siegfried Hornecker [Monday, May 5, 2014 05:45]; edited by Siegfried Hornecker [14-05-05]|
I like Arno's setting as it is now, where even the key has some beauty to it (also the key move reminds me of a famous study that ends with something like wKh8 Ph7 bRh1 Sg8, where White draws with the spectacular 1.Kg7 Rg1+ 2.Kh8 Rh1). I think adding a knight on b1 wouldn't make 1.Sc3 a valid try since after 1.-g6/g5 there are multiple mates.
|(20) Posted by Arno Tungler [Monday, May 5, 2014 06:41]|
Key and matrix are Hauke's, so he is certainly co-author if this turns out to stay the maximum under these conditions (only key + tries - as Rajendiran Raju had requested - plus all tries are refuted by different black moves). As Geir Sune Tallaksen Østmoe had already said, a wSb1 does not even meet the broader requirement as 1.Sc3? is refuted by both bxc3 and Rh6! Anyhow, I am still searching for the 22nd try...
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